SETTLING & SEDIMENTATION IN PARTICLE-

FLUID SEPARATION

• Particles - solid or liquid drops

Settling of a slurry from a soybean leaching process

• fluid - liquid or gas

• Particles are separated from the fluid by gravitation forces

• Applications:

Settling of crystals from the mother liquor

Removal of solids from liquid sewage wastes

Recover particles as the product

Separation of liquid-liquid mixture from a solvent-extraction stage

Remove particles from the fluid (free of particle contaminant)

• Purpose:

Suspend particles in fluids for separation into different sizes or density

MOTION OF PARTICLES THROUGH FLUID

2. buoyant force, which acts parallel with the external force but in the

opposite direction

1. external force, gravitational or centrifugal

3. drag force, which appears whenever there is relative motion between the

particle and the fluid (frictional resistance)

Drag: the force in the direction of flow exerted by the fluid on the solid

Three forces acting on a rigid particle moving in a fluid :

Drag force

External force

Buoyant force

The terminal velocity of a falling object is the velocity of the object when the sum of the drag force (Fd) and buoyancy equals the downward force of gravity (FG) acting on the object. Since the net force on the object is zero, the object has zero acceleration. In fluid dynamics, an object is moving at its terminal velocity if its speed is constant due to the restraining force exerted by the fluid through which it is moving.

Terminal velocity, ut

Drag force

External force, gravity

Buoyant force

Terminal velocity, ut

The terminal velocity of a falling body occurs during free fall when a falling body experiences zero acceleration. This is because of the retarding force known as air resistance. Air resistance exists because air molecules collide into a falling body creating an upward force opposite gravity. This upward force will eventually balance the falling body's weight. It will continue to fall at constant velocity known as the terminal velocity.

Terminal velocity, ut

The terminal velocity of a falling body occurs during free fall when a falling body experiences zero acceleration.

ONE-DIMENSIONAL MOTION OF PARTICLE THRU’

FLUID

a= acceleration of the particle

u = velocity of particle relative to the fluid

where

m = mass of particle

CD = drag coefficient (dimensionless)

Ap = projected area of the particle

FD Drag force Fb Buoyant force

Fe External force

mdudtF

eF

bF

D

Fema

Fbma

p

FDCDu2A

p

2, p= densities of the fluid & particle, respectively

dudtaa

p

CDu2A

p

2ma

pp

CDu2A

p

2m

ONE-DIMENSIONAL MOTION OF PARTICLE THRU’

FLUID

• Motion in a centrifugal field

a = g

where

r = radius of path of particle

= angular velocity, rad/s

• Motion from gravitational force

dudtg

pp

CDu2A

p

2m

a r2

dudt r2

pp

CDu2A

p

2m

u is directed outwardly along a radius

TERMINAL VELOCITY (FREE SETTLING)

when a particle is at a sufficient distance from the walls of the container and

from other particles, so that its fall is not affected by them

• maximum settling velocity (constant velocity) is called terminal/free

settling velocity, ut

ut

2g p

m

AppCD

- period of accelerated fall (1/10 of a second)

- period of constant-velocity fall

Dp = equivalent dia. of particle

where

CD = drag coefficient

, p= densities of the fluid & particle, respectively

g = acceleration of the particle

m = mass of particle

Ap = projected area of the particle

MOTION OF SPHERICAL PARTICLES

Substituting m & Ap into

m 16D

p3p

Ap 1

4D

p2

ut

4g p

D

p

3CD

terminal velocity, ut :

ut

2g p

m

AppCD

DRAG COEFFICIENT FOR RIGID SPHERES

• a function of Reynolds number

restricted conditions:

1) must be a solid sphere particle

2) far from other particles and the vessel wall (flow pattern around the

particle is not distorted)

3) moving at its terminal velocity with respect to the fluid

DRAG COEFFICIENT FOR RIGID SPHERES

DRAG COEFFICIENT

STOKES’ LAW (LAMINAR-FLOW REGION)

= viscosity of fluid (Pa.s or kg/m.s)

Dp = equivalent dia. of particle

applies when NRe 1.0

CD 24N

Re, p

utgD

p2

p

18

FD= total drag force

where CD = drag coefficient

NRe= Reynolds number = (Dput)/

, p= densities of the fluid & particle, respectively

When NRe,p = 1, CD =26.5

NEWTON’S LAW (TURBULENT-FLOW REGION)

CD = 0.44

1000 < NRe,p < 200,000 :

applies to fairly large particles falling in gases or low viscosity fluids

ut1.75

gDpp

Terminal velocity can be found by trial and error by assuming various ut to get

calculated values of CD & NRe which are then plotted on the CD vs NRe

graph to get the intersection on the drag-coefficient correlation line, giving

the actual NRe.

TERMINAL VELOCITY OF A PARTICLE

CRITERION FOR SETTLING REGIME

criterion K :

K Dp

g p

2

1/3

K Region ut

K < 2.6 Stokes’ Law

2.6 < K < 68.9 Intermediate

Region

Trial and Error

68.9 < K < 2360 Newton’s Law

utgD

p2

p

18

ut1.75

gDpp

ut

4g p

D

p

3CD

To determine whether

regime is

Stoke/Intermediate/Newton

TRIAL & ERROR METHOD

criterion K :

K Dp

g p

2

1/3

K Region ut

2.6 < K < 68.9 Intermediate

Region

ut

4g p

D

p

3CD

Terminal velocity can be found by trial and error by:

Step 1: Assume NRe which then will give CD from the CD vs NRe graph.

Step 2: Calculate ut.

Step 3: Using the calculated ut, the NRe is checked to verify if it agrees with the

assumed value.

DRAG COEFFICIENT

Example 1

Solid spherical particles of coffee extract from a dryer having a diameter

of 400 m are falling through air at a temperature of 422 K. The density

of the particles is 1030 kg/m3. Calculate the terminal settling velocity and

the distance of fall in 5 s. The pressure is 101.32 kPa.

K Dp

g p

2

1/3

Example 2

Oil droplets having a diameter of 20 mm are to be settled from air at

311K and 101.3 kPa pressure. The density of the oil droplets is 900

kg/m3. Calculate the terminal settling velocity of the droplets.

K Dp

g p

2

1/3

Physical Properties of Air

HINDERED SETTLING

• uniform suspension

equation of Maude & Whitmore us = ut ( ε )n

where us = settling velocity

ut = terminal velocity for an isolated particle

= total void fraction (fluid fraction)

• velocity gradients around each particle are affected

by the presence of nearby particles

• particle velocity relative to the fluid > the absolute

settling velocity

• large number of particles are present

n = exponent n from figure 7.8 (page 52 course

notes)

HINDERED SETTLING

• Suspensions of very fine sand in water :

• larger particles thru’ a suspension of much finer solids:

ε = volume fraction of the fine suspension, not the total void fraction

ut calculated using the density and viscosity of the fine suspension

us = ut ( ε )n

used in separating coal from heavy minerals

density of the suspension is adjusted to a value slightly greater than

that of coal to make the coal particles rise to the surface, while the

mineral particles sink to the bottom

HINDERED SETTLING

us = ut ( ε )n

Example 3

1. (a) Estimate the terminal velocity for 80-to-100 mesh

particles of limestone (p = 2800 kg/m3) falling in water at

30oC.

(b) How much higher would the velocity be in a centrifugal

separator where the acceleration is 50g?

* Refer to page 57 for properties of water

Physical Properties of Water

• Free Settling – is when the fall of a particle is not affected by the boundaries of the container and from other particles (due to a sufficient distance between the particle-container and particle - particle).

• Hindered Settling – is when the fall is impeded by other particles because the particles are near to another.

• CD hindered settling > CD free settling

Flow past immersed bodies Slide28

Hindered Settling

• In hindered settling, the velocity gradients around each particle are effected by the presence of nearby particles; so the normal drag correlations do not apply.

• Furthermore, the particles in settling displace liquid, which flows upward and make the particle velocity relative to the fluid greater than the absolute settling velocity, us.

• For uniform suspension, the settling velocity can be estimated from the terminal velocity for an isolated particle using the empirical equation of Maude and Whitmore :

us = ut ()n

where is a total void fraction.

Flow past immersed bodies Slide29

----- Eq 7.46

Solution • SG = p/ SG = p SG- = p- (SG-1) = p-

• Dp = 0.004 in = 0.004/12 ft

• 1 cP = 6.7197 x 10-4 lb/ft.s

• g = 32.174 ft/s2

• Use Eq 7.40 to find ut:

• Calculate Rep using Eq 7.44:

• Use Rep value to find exponent n from Fig 7.8.

• Use Eq 7.46 to find ut in hindered settling

Flow past immersed bodies Slide30

ut gDp

2 p 18

us = ut ()n

tp

ep

uDR

VO

FLUIDIZATION

fluid is passed at a very low velocity up through a bed of solid, particles do

not move (fixed bed)

At high enough velocity fluid drag plus buoyancy overcomes the gravity force so

particle start to move/suspended and the bed expands (Fluidized Bed).

FLUIDIZATION

2 types of fluidization:

(2) bubbling fluidization - bubbles with only a small % of gas passes in the

spaces between particles, little contact between bubbles & particles

(1) particulate fluidization - bed remains homogeneous, intimate contact

between gas & solid

1 2 3 4 5 6

FLUIDIZATION

fully suspended particles & bed expands ( the suspension behave like a

dense fluid ).

fluidized solids can be drained from the bed through pipes and valves just as a

liquid can

Applications:

Fluidized bed drying

Fluidized bed combustion

Fluidized bed reactions

FLUIDIZATION

L is constant until onset of fluidization and then begins to increase.

Until onset of fluidization p increases, then becomes constant.

MINIMUM FLUIDIZATION VELOCITY

where

At the point of incipient/beginning fluidization :

pressure drop across the bed equal to the weight of the bed per unit area :

pressure drop given by Ergun Eq :

PL

150Vo

gs2D

p2

1

2

31.75V o

2

gsDp

13

Pg1

p

L

150VOMs2D

p2

1M

M3

1.75VOM

2

sDp

1M3g

p

= minimum fluidization velocity (fluid vel. at which fluidization begins)

VOM

= minimum bed porosity/void fraction

M

L

S

MINIMUM FLUIDIZATION VELOCITY

At the point of incipient/beginning fluidization :

150VOMs2D

p2

1M

M3

1.75VOM

2

sDp

1M3g

p

Void Fraction at Min. Fluidization

M depends on the shape of the particles. For spherical particles M is

usually 0.4 – 0.45.

Example 4

A bed of ion-exchange beads 8 ft deep is to be backwashed with water to

remove dirt. The particle have a density of 1.24g/cm3 and an average

size of 1.1 mm. What is the minimum fluidization velocity using water at

20oC. The beads are assumed to be spherical ( = 1 ) and is taken as

0.4.

M

s

150VOMs2D

p2

1M

M3

1.75VOM

2

sDp

1M3g

p

MINIMUM FLUIDIZATION VELOCITY USING NRE

At the point of incipient/beginning fluidization :

Minimum fluidization Reynolds number :

In term of minimum fluidization Reynolds number:

150VOMs2D

p2

1M

M3

1.75VOM2

sDp

1M3g

p

NReM

DpVOM

150 1M

(N

ReM)

s2M3

1.75(N

ReM)2

sM3

Dp

3g p 2

MINIMUM FLUIDIZATION VELOCITY

ratio of ut/VOM :

• very small particles (NRe,p <1) :

only the laminar-term is significant

VOM

g

p

150M3

1M

s2D

p2

ut

VOMgD

p2

p

18150

g p

s2D

p2

1M

M3

8.331

M

s2M3

MINIMUM FLUIDIZATION VELOCITY

• Larger Particles (NRe,p > 1000, larger than 1 mm) :

ratio of ut/VOM :

VOM sDpg

p

M3

1.75

1/2

ut

VOM1.75

gDpp

1/2

1.75

gDpp

M3

1/2

2.32M3/2

ut = terminal settling velocity of the particles ( maximum allowable velocity)

MINIMUM FLUIDIZATION VELOCITY

Substituting into the minimum fluidization velocity eq. :

If M & S are unknown:

Reasonable estimate ( 25%)

SM

3 114

1M

S

2M

311

NReM

(33.7)20.0408gD

p

3

p

2

1/ 2

33.7

Holds for 0.001 Nre < 4000

BED LENGTH AT MINIMUM FLUIDIZATION

S = cross-sectional area of fluidized bed

LM = minimum bed height at onset of fluidization

M = void fraction at minimum fluidization

Bed height is needed in order to size the vessel

where

LM mS 1

M

p

LM

S

p = density of particle

m = mass of particles

EXPANSION OF FLUIDISED BEDS

= void fraction at operating velocity

Particulate fluidization

where

LLM

1M

1

L = expanded bed height

Small particles & NRe,p 20 :

VODp

2g p

S

2

1503

1K

1

3

1

= operating velocity

VO

L

S

VO

Example 5

Solid particles having a size of 0.12 mm, a shape factor of 0.88, and a density

of 1000 kg/m3 are to be fluidized using air at 2 atm abs and 25oC. The voidage at

minimum fluidizing conditions is 0.42.

a. If the cross section of the empty bed is 0.3 m2 and the bed contains 300 kg of

solid, calculate the minimum height of the fluidized bed.

b. Calculate the presure drop at minimum fluidizing conditions.

c. Calculate the minimum velocity for fluidization.

d. Assuming that data for Φs and εm are unavailable, calculate the minimum

fluidization velocity

M

s

Example 5

150VOMs2D

p2

1M

M3

1.75VOM

2

sDp

1M3g

p

Pg1

p

L

LM mS 1

M

p

NReM

(33.7)20.0408gD

p

3

p

2

1/ 2

33.7